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(·)η : ̺ → ̺∗ is a ∧-homomorphism of L(S) into itself. In Theorem 3.1, we show that ̺ is a left equalizer simple congruence on a semigroup S if and only if the congruence ̺∗ is a left cancellative congruence on S. Let S be a semigroup and F a field. For an arbitrary congruence ̺ on S, let F[̺] denote the ideal of the semigroup algebra F[S] which determines the kernel of the extended homomorphism of the semigroup algebra F[S] onto the semigroup algebra F[S/̺] defined by the canonical homomorphism of S onto the factor semigroup S/̺ (see [1]). For an ideal J of F[S], let J|S denote the restriction of the congruence on F[S] defined by J to S. By Lemma 5 of Chapter 4 of [10], for every semigroup S and every field F, the mapping J 7→ J|S is a surjective homomorphism of the ∧-semilattice Id(F[S]) of all ideals of F[S] onto the ∧-semilattice of all congruences on S such that F[̺]|S = ̺ for every congruence ̺ on S. Let (·)κ : ̺ → F[̺]. By the above, κ is an injective mapping of L(S) into Id(F[S]). Using the notation of Section 3.6 of [6] for semigroup algebras, if J is an arbitrary ideal of F[S] then let (J :r F[S]) = {a ∈ F[S]: F[S]a ⊆ J}. It is easy to see that (J :r F[S]) is an ideal of the semigroup algebra F[S] such that J ⊆ (J :r F[S]). An ideal (J :r F[S]) will be called a right colon (more precisely, the right colon of J with respect to F[S]). It is a matter of checking to see that (J1 ∩ J2 :r F[S]) = (J1 :r F[S]) ∩ (J2 :r F[S]) for arbitrary ideals J1 and J2 of F[S]. Thus (·)Φ : J → (J :r F[S]) is a ∧-homomorphism of Id(F[S]) into itself. By the above, we can consider the following diagram.

L(S) −→η L(S) κ ↓ ↓ κ (1) −→ Id(F[S]) Φ Id(F(S])

2 We shall say that the diagram (1) is commutative for some congruence ̺ on a semigroup S if (̺)(η ◦ κ) = (̺)(κ ◦ Φ), that is, ∗ F[̺ ] = (F[̺]:r F[S]). In Theorem 3.2, we prove that, for arbitrary left equalizer simple congruence on a semigroup S, the diagram (1) is commutative. In Section 4, our results will be applied for the matrix representation of finite left equalizer simple semigroups. We also prove a theorem for arbitrary (not necessarily finite) semigroups. We show that if ̺ is a left equalizer simple congruence on a semigroup S then the right colon (F[̺]:r F[S]) equals the augmentation ideal F[ωS] of the semigroup algebra F[S] if and only if the factor semigroup S/̺ is an ideal extension of a left zero semigroup by a null semigroup.

2 Left equalizer simple semigroups

Let S be a semigroup and H a non-empty subset of S. By the left equalizer of H we mean the set of all elements x of S for which |xH| = 1, that is, xa = xb is satisfied for all a,b ∈ H. It is clear that the left equalizer of H is either empty or a left ideal of S.

Lemma 2.1 On an arbitrary semigroup S, the following assertions are equiv- alent. (i) The left equalizer of any two-element subset of S is either empty or S. (ii) The left equalizer of any subset of S is either empty or S. Proof. Assume (i). Let H be a non-empty subset of S. If |H| = 1 then the left equalizer of H is S. Consider the case when |H| ≥ 2. If x0 ∈ S is in the left equalizer of H then, for every two elements a,b ∈ H with a 6= b, we have x0a = x0b and so x0 belongs to the left equalizer of the subset {a,b}. Thus every x ∈ S is in the left equalizer of {a,b}, that is, xa = xb for all x ∈ S. As a,b ∈ H are arbitrary, we have |xH| = 1 for all x ∈ S and so the left equalizer of H is S. It is obvious that condition (ii) implies condition (i). ⊓

Definition 2.1 A semigroup S will be called a left equalizer simple semigroup if, for arbitrary non-empty subset H of S, the left equalizer of H is either empty or equal to S. Equivalently (see Lemma 2.1), for arbitrary elements a,b ∈ S, the assumption x0a = x0b for some x0 ∈ S implies xa = xb for all x ∈ S.

A semigroup S is said to be left simple if S is the only left ideal of S. It is obvious that every left simple semigroup is left equalizer simple.

3 Lemma 2.2 Every M-inversive semigroup is left equalizer simple.

Proof. Let x0,a,b be arbitrary elements of an M-inversive semigroup S with x0a = x0b. Then there is an element y ∈ S such that yx0 is a middle unit of S and so, for all x ∈ S, we have xa = xyx0a = xyx0b = xb. Hence S is left equalizer simple. ⊓ The next theorem is an extension of Theorem 1 of [4].

Theorem 2.1 A semigroup S is left equalizer simple if and only if the right regular representation of S is a left cancellative semigroup. Proof. Let S be a left equalizer simple semigroup. Let ϕ denote the canon- ical homomorphism of S onto the right regular representation of S. If

ϕ(x0)ϕ(a)= ϕ(x0)ϕ(b) for some x0,a,b ∈ S then, for an arbitrary element x1 ∈ S, we have

x1x0a = x1x0b.

As S is left equalizer simple, it follows that xa = xb for all x ∈ S. Hence ϕ(a) = ϕ(b). Thus the right regular representation of S is a left cancellative semigroup. Conversely, assume that the right regular representation ϕ(S) of a semigroup S is left cancellative. If x0a = x0b for some x0,a,b ∈ S then

ϕ(x0)ϕ(a)= ϕ(x0)ϕ(b) in ϕ(S). As ϕ(S) is left cancellative, we get ϕ(a) = ϕ(b) which means that xa = xb for all x ∈ S. Thus S is a left equalizer simple semigroup. ⊓ Construction 1: Let T be a left cancellative semigroup. For each t ∈ T , associate a nonempty set St such that St ∩ Sr = ∅ for every t 6= r. As T is left cancellative, x 7→ tx is an injective mapping of T onto tT . For arbitrary couple (t, r) ∈ T × T with r ∈ tT , let f(t,r) be a mapping of St into Sr. For all t ∈ T , r ∈ tT , q ∈ rT ⊆ tT and a ∈ St, assume

(a)f(t,r) ◦ f(r,q) = (a)f(t,q).

On the set S = ∪t∈T St define an operation ⋆ as follows: for arbitrary a ∈ St and b ∈ Sx, let a⋆b = (a)f(t,tx).

If a ∈ St, b ∈ Sx, c ∈ Sy are arbitrary elements then

a⋆ (b⋆c)= a⋆ (b)f(x,xy) = (a)f(t,t(xy)) =

= (a)(f(t,tx) ◦ ftx,t(xy) = (a)f(t,tx) ⋆c = (a⋆b) ⋆ c. Thus (S; ⋆) is a semigroup.

4 Let x ∈ T and a ∈ Sx be arbitrary. Then, for arbitrary b ∈ Sx and c ∈ St (t ∈ T ), we have c⋆a = (c)f(t,tx) = c ⋆ b.

Thus (a,b) ∈ θS and so Sx is contained by the θS-class [a]θS of S. If d ∈ [a]θS ∩Sy (y ∈ T ) then, for an arbitrary c ∈ St (t ∈ T ), we have

(c)f(t,tx) = c⋆a = c ⋆ d = (c)f(t,ty) from which we get tx = ty. As T is left cancellative, we get x = y and so d ∈ Sx.

Consequently Sx = [a]θS . Hence each set Sx (x ∈ T ) is a θS-class of S.

Example 2.1 Let T = {1, 2} be a two-element group in which 1 is the identity element. Let S1 = {x1,y1} and S2 = {x2,y2} be disjoint sets. For arbitrary t,s ∈ T , let f(t,ts) defined by: (xt)f(t,ts) = xts and (yt)f(t,ts) = yts. It is easy to see that the conditions of Construction 1 are satisfied. The Cayley-table of the semigroup (S; ⋆) is Table 1.

x1 y1 x2 y2 x1 x1 x1 x2 x2 y1 y1 y1 y2 y2 x2 x2 x2 x1 x1 y2 y2 y2 y1 y1

Table 1:

Theorem 2.2 A semigroup is a left equalizer simple semigroup if and only if it is isomorphic to a semigroup defined in Construction 1. Proof. Let (S; ⋆) be a semigroup defined in Construction 1. We show that (S; ⋆) is left equalizer simple. Assume

x0 ⋆a = x0 ⋆b for some x0,a,b ∈ S. Let ξ,t,r ∈ T such element for which x0 ∈ Sξ, a ∈ St and b ∈ Sr. Then (x0)f(ξ,ξt) = (x0)f(ξ,ξr) and so ξt = ξr in T . As T is left cancellative, t = r and so ηt = ηr for all η ∈ T . Thus, for every η ∈ T and x ∈ Sη,

x⋆a = (x)f(η,ηt) = (x)f(η,ηr) = x ⋆ b.

Consequently S is a left equalizer simple semigroup.

5 To prove the converse, let S be a left equalizer simple semigroup. Then, by Theorem 2.1, the right regular representation T = ϕ(S) = S/θS is a left cancellative semigroup. Construct the semigroup as in Construction 1 defined by the semigroup T = S/θS. For arbitrary t ∈ T , let St be the θS-class of S for which ϕ(St) = t. For arbitrary couple (t, r) ∈ T × T with r ∈ tT , let f(t,r) be a mapping of St into Sr defined by the following way. As T is left cancellative, there is a unique element x ∈ T such that tx = r. For arbitrary a ∈ St, let (a)f(t,r) = aw, where w is an arbitrary element of Sx. For arbitrary a ∈ St and b ∈ Sx, a⋆b = (a)f(t,tx) = ab. Hence, S is isomorphic to the semigroup (S; ⋆). ⊓

3 Left equalizer simple congruences

According to our definition for the P congruence from Section 1, a congruence ̺ on a semigroup S is a left equalizer simple congruence if, for every a,b ∈ S, the assumption (x0a, x0b) ∈ ̺ for some x0 ∈ S implies (xa, xb) ∈ ̺ for all x ∈ S. A congruence ̺ on a semigroup S is a left cancellative congruence if, for every a,b ∈ S, the assumption (x0a, x0b) ∈ ̺ for some x0 ∈ S implies (a,b) ∈ ̺.

Theorem 3.1 A congruence ̺ on a semigroup S is a left equalizer simple con- gruence if and only if ̺∗ is a left cancellative congruence on S. Proof. Let ̺ be an arbitrary congruence on a semigroup S. As ̺ ⊆ ̺∗, we can consider the congruence ̺∗/̺ (see [12]). By Theorem 5.6 of [12], ∗ ∼ ∗ ∗ ∗ (S/̺)/(̺ /̺) = S/̺ . By Lemma 7 of [7], ̺ /̺ = ιS/̺, where ιS/̺ denotes ∗ the identity relation on the factor semigroup S/̺. As ιS/̺ = θS/̺, we get that the semigroup S/̺∗ is the right regular representation of S/̺. Thus ̺ is a left equalizer simple congruence on S if and only if the factor semigroup S/̺ is a left equalizer simple semigroup. By Theorem 2.1, this last condition is equivalent to the condition that the right regular representation of S/̺, that is, the semigroup S/̺∗ is left cancellative. This last condition means that ̺∗ is a left cancellative congruence on S. ⊓

Let S be a semigroup and F a field. Let F[S] denote the semigroup algebra of S over F. By page 159 of [1], S can be considered as the basis of F[S] and F every element a of [S] can be written as a sum s∈S a(s)s; this is a finite sum since only a finite number of coefficients a(s) ∈ F are 6= 0. For arbitrary elements a,b ∈ F[S] and α ∈ F, P

a + b = (a(s)+ b(s))s, Xs∈S

ab = a(x)b(y) s,   Xs∈S {x,y∈XS;xy=s}   6 αa = (α(a(s))) s. Xs∈S F F It is easy to see that an element a = s∈S a(s)s of [S] is in [̺] if and only if s∈A a(s) = 0 for every ̺-class A of S. If ωS denotes the universal relation F P F on S then the ideal [ωS] is the set of all elements s∈S a(s)s of [S] for which P a(s) = 0; this ideal is the (so called) augmentation ideal of F[S]. s∈S P LemmaP 3.1 For an arbitrary field F and an arbitrary congruence ̺ on a semi- ∗ group S, F[̺] ⊆ F[̺ ] ⊆ (F[̺]:r F[S]) (with the notations of the diagram (1), (̺)κ ⊆ (̺)(η ◦ κ) ⊆ (̺)(κ ◦ Φ)). Proof. The algebra ideal F[α] of a semigroup congruence α is obviously generated (as an F-space) by differences s − s′, where (s,s′) ∈ α. Thus the inclusion F[̺] ⊆ F[̺∗] is obvious. If (s,s′) ∈ ̺∗ then, for every x ∈ S, (xs, xs′) ∈ ̺, hence xs − xs′ ∈ F[̺]. ∗ From this it follows that F[̺ ] ⊆ (F[̺]:r F[S]). ⊓

Example 3.1 Let S = {a,b,c,d,e} be a semigroup defined by Table 2 (see [2]; page 167, the last Cayley-table in row 7):

abc de a b baab b b bb bb c b bccb d b bddb e abab e

Table 2:

As the columns of the Cayley-table are pairwise distinct, S is left reductive and so the identity relation ιS on S is left reductive. Thus, by Theorem 1 of [7],

∗ ιS = ιS and so F F ∗ [ιS]= {0} = [ιS ]. It is a matter of checking to see that

d + a − b − c ∈ (F[ιS]:r F[S]).

Thus {0}⊂ (F[ιS ]:r F[S]). ∗ This example shows that F[̺] = (F[̺]:r F[S]) or F[̺ ] = (F[̺]:r F[S]) is not satisfied for an arbitrary congruence ̺ on a semigroup S.

7 By Lemma 3.1, for an arbitrary field F and an arbitrary congruence ̺ on a semigroup S: (a) F[̺] ⊆ F[̺∗],

(b) F[̺] ⊆ (F[̺]:r F[S]),

∗ (c) F[̺ ] ⊆ (F[̺]:r F[S]).

In case (a), the equation F[̺] = F[̺∗] holds if and only if ̺ = ̺∗, because the mapping κ : ̺ → F[̺] is injective (by Lemma 5 of Chapter 4 of [10]). By Theorem 1 of [7], a congruence ̺ on a semigroup S is left reductive if and only if ̺ = ̺∗. Thus we have the following proposition.

Proposition 3.1 Let F be an arbitrary field. The equation F[̺] = F[̺∗] is satisfied for a congruence ̺ on a semigroup S if and only if ̺ is a left reductive congruence on S. ⊓

To characterize congruences ̺ on a semigroup S for which the equation F[̺] = (F[̺]:r F[S]) holds (see case (b)), consider the notion of the right annihi- lator Annr(F[S]) of a semigroup algebra F[S]. Recall that Annr(F[S]) = {a ∈ F[S] : (∀x ∈ F[S]) xa = 0} = ({0} :r F[S]). The right annihilator is said to be trivial if it contains only the zero 0 of F[S]. Here we refer to [8] and [9], where the finite semigroups S with condition Annr(F[S]) = {0} were examined.

Proposition 3.2 Let F be an arbitrary field. For a congruence ̺ on a semi- group S, the equation F[̺] = (F[̺]:r F[S]) is satisfied if and only if the right annihilator Annr(F[S/̺]) is trivial. Proof. Let τ denote the (extended) canonical homomorphism of F[S] onto F[S/̺]. For an element a ∈ F[S], τ(a) ∈ Annr(F[S/̺]) if and only if, for all x ∈ F[S], τ(xa) = τ(x)τ(a) = 0 which is equivalent to the condition that xa ∈ F[̺] for all x ∈ F[S]. This last condition means that a ∈ (F[̺]:r F[S]). From this it follows that F[̺] = (F[̺]:r F[S]) if and only if Annr(F[S/̺]) is trivial. ⊓ ∗ The condition that equation F[̺ ] = (F[̺]:r F[S]) holds in case (c) is equiv- alent to the condition that the diagram (1) is commutative for ̺. The next theorem shows that the left equalizer simplicity is a sufficient condition for a congruence ̺ to be the diagram (1) commutative for ̺.

Theorem 3.2 Let F be an arbitrary field. If ̺ is a left equalizer simple congru- ∗ ence on a semigroup S then F[̺ ] = (F[̺]:r F[S]), that is, (̺)(η ◦κ) = (̺)(κ◦Φ) which means that the diagram (1) is commutative for ̺. Proof. Let ̺ be a left equalizer simple congruence on a semigroup. We must show that ∗ F[̺ ] = (F[̺]:r F[S]).

8 By Lemma 3.1, it is sufficient to show that

∗ (F[̺]:r F[S]) ⊆ F[̺ ].

Let a(s)s ∈ (F[̺]:r F[S]) Xs∈S be arbitrary. Let x ∈ S be an arbitrary element. Then

a(s)xs ∈ F[̺]. Xs∈S Let A be a ̺∗-class of S, and let B denote the ̺-class of S containing xA. We show that xS ∩ B = xA. The inclusion xA ⊆ xS ∩ B is obvious. If xs ∈ B for some s ∈ S then (xs, xa) ∈ ̺ for some a ∈ A. As ̺ is a left equalizer simple congruence on S, we get (ts,ta) ∈ ̺ for every t ∈ S. Thus (s,a) ∈ ̺∗ and so s ∈ A. This implies xS ∩ B ⊆ xA. Consequently xS ∩ B = xA. Thus the sum s∈A a(s) is the same as the sum xs∈B a(s). The latter sum is 0 by assumption, giving that the former sum is also 0, which proves the theorem. ⊓ P P

4 Matrix representations

Let S be a finite semigroup and F an arbitrary field. By an S-matrix over F we mean a mapping of the direct product S × S into F. The set FS×S of all S- matrices over F is an algebra over F under the usual addition and multiplication of matrices and the product of matrices by scalars. A homomorphism γ of a S into the multiplicative semigroup of the full matrix algebra Fn×n of all n × n matrices over F is called a matrix representation of S of order n. We say that γ is faithful if it is injective. For an element s ∈ S, let R(s) denote the S-matrix defined by

1, if xs = y R(s)((x, y)) = (0 otherwise, where 1 and 0 denote the identity element and the zero element of F, respectively. This matrix is called the right matrix of the element s of S. The mapping

(s) RF : s 7→ R is a matrix representation of S over F (see, for example, Exercise 4 in §3.5 of [1]). RF describes in terms of matrices R(s) the maps x 7→ xs. Thus, it is essentially the right regular representation. It is obvious that a semigroup containing an identity element is left reductive. Thus, for an arbitrary finite semigroup S, the restriction R′F of the right regular matrix representation of S1 to S is a faithful matrix representation of S. The matrices {R′F(s); s ∈ S} are also linearly independent over F.

9 If S is an arbitrary finite n-element left reductive semigroup then the sys- tem {RF(s); s ∈ S} has n pairwise different matrices, but these matrices are not linearly independent over F, in general. By Definition 2.2 of [9], a finite semi- group S is called an RF-independent semigroup if the system {RF(s); s ∈ S} of matrices is linearly independent over F. For a congruence ̺ on S, define the following sequence (see the diagram (1)):

̺(n) = (̺)ηn, n =0, 1,...

If ̺ is a left equalizer simple congruence on S then ̺(1) is a left cancellative congruence on S by Theorem 3.1. As a left cancellative congruence is also left reductive, we have ̺(1) = ̺(2) = ··· by Theorem 1 and Theorem 2 of [7]. As a left cancellative congruence is also left equalizer simple, we have

(2) (1) F[̺ ] = (F[̺ ]:r F[S]) by Theorem 3.2. From this it follows that

(1) (1) F[̺ ] = (F[̺ ]:r F[S]). (2)

(1) By Proposition 3.2, it means that Annr(F[S/̺ ]) is trivial. Then the factor semigroup S/̺(1) is RF-independent by Theorem 2.1 of [9]. The above result is not too interesting if S/̺(1) is a one-element semigroup, (1) that is, if ̺ = ωS (ωS denotes the universal relation on S). This special case is equivalent to (see also (2))

(1) (1) (F[̺ ]:r F[S]) = F[̺ ]= F[ωS], where F[ωS] is the augmentation ideal of F[S]. In the last part of the paper we deal with the following question: What can we say about the factor semigroup S/̺ if ̺ is a left equalizer simple congruence on a (not necessarily finite) semi- group S such that the right colon (F[̺]:r F[S]) equals the augmentation ideal F[ωS] of F[S].

Theorem 4.1 Let ̺ be a left equalizer simple congruence on a semigroup S and F a field. The right colon (F[̺]:r F[S]) equals the augmentation ideal F[ωS] of F[S] if and only if the factor semigroup S/̺ is an ideal extension of a left zero semigroup by a null semigroup (which means that S has an ideal L which is a left zero semigroup and the Rees factor semigroup S/L is a null semigroup (which means that (S/L)2 = {0})).

(1) Proof. As ̺ is left equalizer simple, F[̺ ] = (F[̺]:r F[S]) by Theorem 3.2. (1) Thus (F[̺]:r F[S]) = F[ωS] is satisfied if and only if F[̺ ] = F[ωS]. By (1) Lemma 5 of Chapter 4 of [10], this is equivalent to ̺ = ωS. By the proofs of Theorem 5 and Theorem 6 of [7], this is equivalent to the condition that the

10 factor semigroup S/̺ is an ideal extension of a left zero semigroup by a null semigroup. ⊓

In the next example we give a semigroup S in which the identity relation ιS is a left equalizer simple congruence on S such that the right colon of F[ιS] with respect to F[S] equals the augmentation ideal of F[S].

Example 4.1 Let S = {a,b,c,d} be a semigroup defined by Table 3.

ab c d a a a a a b b b b b c a a a a d b b b b

Table 3:

It is easy to see that S can be obtained by applying Construction 1 in that case when T = {e} is a one-element semigroup, Se = {a,b,c,d} and f(e,e) defined by

(a)f(e,e) = a, (b)f(e,e) = b, (c)f(e,e) = a, (d)f(e,e) = b.

It is a matter of checking to see that

f(e,e) ◦ f(e,e) = f(e,e) and so the conditions of Construction 1 are satisfied. The semigroup S is left equalizer simple and so the identity relation ιS of S is a left equalizer simple congruence on S. For an element αa + βb + γc + δd ∈ F[S],

F (1) F F αa + βb + γc + δd ∈ [ιS ] = ( [ιS]:r [S]) if and only if (α + β + γ + δ)=0.

Thus the right colon (F[ιS]:r F[S]) equals the augmentation ideal F[ωS]. It is easy to see that S is an ideal extension of the left zero semigroup L = {a,b} by the null semigroup S/L.

Acknowledgement: I would like to thank the referee for the valuable re- marks on the original version of the paper.

11 References

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